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Probability of finding a particle at an energy state is:

$\mathbf{P}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{=}\frac{\mathbf{(}\mathbf{M}\mathbf{鈭�}\mathbf{n}\mathbf{+}\mathbf{N}\mathbf{鈭�}\mathbf{2}\mathbf{)}\mathbf{!}}{\mathbf{(}\mathbf{M}\mathbf{鈭�}\mathbf{n}\mathbf{)}\mathbf{!}\mathbf{(}\mathbf{N}\mathbf{鈭�}\mathbf{2}\mathbf{)}\mathbf{!}}\mathbf{/}\frac{\mathbf{(}\mathbf{M}\mathbf{+}\mathbf{N}\mathbf{鈭�}\mathbf{1}\mathbf{)}\mathbf{!}}{\mathbf{M}\mathbf{!}\mathbf{(}\mathbf{N}\mathbf{鈭�}\mathbf{1}\mathbf{)}\mathbf{!}}$ 鈥︹��. (1)

where$\mathbf{M}$, is the sum of nonnegative distinct quantum numbers${\mathbf{n}}_{\mathbf{i}}$and$\mathbf{N}$is the total number of particles.

Here, $\mathrm{M}=2,\mathrm{N}=4$, and the allowed values of $\mathrm{n}$are $\mathrm{n}=0,1,2$. Let us first consider $\mathrm{n}=0$.

The probability is:

role="math" localid="1660022722768" $\begin{array}{l}\mathrm{P}(\mathrm{n}=0)=\frac{(2鈭�0+4鈭�2)!}{(2鈭�0)!(4鈭�2)!}/\frac{(2+4鈭�1)!}{2!(4鈭�1)!}\\ =\frac{4!}{2!2!}/\frac{5!}{2!3!}\\ =6/10\\ =0.6\end{array}$

For$\mathrm{n}=1$, we get:

$\begin{array}{l}\mathrm{P}(\mathrm{n}=1)=\frac{(2鈭�1+4鈭�2)!}{(2鈭�1)!(4鈭�2)!}/\frac{(2+4鈭�1)!}{2!(4鈭�1)!}\\ =\frac{3!}{1!2!}/\frac{5!}{2!3!}\\ =3/10\\ =0.3\end{array}$

Finally, for $\mathrm{n}=2$, we get:

$\begin{array}{l}\mathrm{P}(\mathrm{n}=2)=\frac{(2鈭�2+4鈭�2)!}{(2鈭�2)!(4鈭�2)!}/\frac{(2+4鈭�1)!}{2!(4鈭�1)!}\\ =\frac{2!}{0!2!}/\frac{5!}{2!3!}\\ =1/10\\ =0.1\end{array}$

We have thus verified that the probabilities agree with the result from equation 1.